nLab orthogonal group of an inner product space

Idea

Given an inner product space $V = (V,\langle-,-\rangle)$, the orthogonal group of $V$ is the subgroup of the general linear group $GL(V)$ which leaves invariant the inner product.

Definition

Given an element $A$ of $GL(V)$ we say it preserves the inner product $\langle-,-\rangle$ if $\langle A v ,A w \rangle = \langle v,w \rangle$ for all $v,w\in V$.

Proposition

If $A$ and $B$ preserve the inner product on $V$, then so do $AB$ and $A^{-1}$.

We thus see that we have a group.

Definition

The orthogonal group $O(V,\langle-,-\rangle)$ is the subgroup of $GL(V)$ consisting of those elements that preserve the inner product.

(… more detail …)

As a Lie group

When the base field is $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$ (the last being a skewfield, but this is ok), and the vector space is finite-dimensional then we can put the structure of a finite-dimensional manifold on $O(V,\langle-,-\rangle)$. We shall denote the group in this case by $O(n,\mathbb{K})$ where $V = \mathbb{K}^n$, $\mathbb{K} \in \{ \mathbb{R},\mathbb{C},\mathbb{H}\}$ and we take the ‘standard’ inner products on these spaces.

A relatively easy way to see that $O(n,\mathbb{K})$ is a manifold is that it is a smooth affine variety in Euclidean space, but this requires some machinery. We can however construct explicit charts for $O(n,\mathbb{K})$ as a real manifold.

Charts on the underlying manifold

One can use the algebra structure on matrices over the base field (or base division ring) of $V$ to define charts which are different to the charts one gets via the exponential map.

For $A \in O(n,\mathbb{K})$ Define the sets

$$\Omega(A) := \{ X\in O(n,\mathbb{K}) | A + X \in GL(V) \}$$

and

$$T_{A^\dagger} := \{ Y\in M_n(\mathbb{K}) | AY + (AY)^\dagger = 0 \}.$$

Here $T_{A^\dagger}$ is a vector space, and will turn out to be the tangent space to $O(n,\mathbb{K})$ at $A^\dagger$.

Proposition

There is an isomorphism $c_A\colon \Omega(A) \stackrel{\sim}{\to} T_{A^\dagger}$.

Proof

The map $c_A$ is defined as

$$X \mapsto (I - A^\dagger X)(A + X)^{-1} = (I - A^\dagger X)(I + A^\dagger X)^{-1}A^\dagger$$

the latter equality using that $A^{-1} = A^\dagger$. We need to check that this is indeed a map to $T_{A^\dagger}$ and that it is an isomorphism.

Since

$$(A c_A(X))^\dagger = A c_I(A^\dagger X)^\dagger A^\dagger,$$

if we show that $c_I(A^\dagger X)^\dagger = - c_I(A^\dagger X)$ we are done. Let $Z = X^\dagger A$, which is again in $O(n,\mathbb{K})$. Then

$$\begin{aligned} c_I(Z)^\dagger & = \left( (I - Z)(I + Z)^{-1} \right)^\dagger\\ & = \left( (I - Z)Z^\dagger Z^{\dagger -1}(I + Z)^{-1} \right)^\dagger\\ & = \left( -(I - Z^\dagger)(I + Z^\dagger)^{-1} \right)^\dagger \\ & = -(I + Z)^{-1}(I - Z) \end{aligned}$$

(tbc…)

Examples

• Euclidean space$\mathbb{R}^n$ with the standard inner product $\langle\mathbf{x},\mathbf{y}\rangle = \mathbf{x}\cdot\mathbf{y}$ has as orthogonal group the standard orthogonal group $O(n)$.

• The finite-dimensional Hilbert space $\mathbb{C}^n$ with the standard inner product has as orthogonal group $U(n)$, the unitary group

• The space $\mathbb{H}^n$, for $\mathbb{H}$ the quaternions, has an inner product … such that the corresponding orthogonal group is the compact symplectic group $Sp(n)$.

• (…Examples of indefinite signature go here…)